Gaussian Laws and Covariances of Deterministic Wiener Integrals

From Individual Integrals to a Joint Law

The mean and variance of one deterministic Wiener integral do not by themselves describe how several such integrals vary together. If the same Brownian motion drives them, their dependence is determined by how their integrands overlap: the covariance of two integrals is the $L^2$ inner product of those integrands. This gives a direct route from a finite list of functions to a joint Gaussian law, and it also identifies when the integrals are independent or linearly dependent.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to the filtration. Fix $T>0$. For each Borel measurable deterministic function $f:[0,T]\to\mathbb{R}$ with $\int_0^T f(t)^2\,dt<\infty$, the process $H_t=f(t)$ is adapted and progressively measurable. Indeed, its restriction to each $[0,s]\times\Omega$ is measurable with respect to $\mathcal{B}([0,s])\otimes\mathcal{F}_s$. Also, $\mathbb{E}\int_0^T H_t^2\,dt=\int_0^T f(t)^2\,dt<\infty$. Thus $H$ belongs to the class $\mathcal{L}^2$ of progressively measurable integrands with finite expected integrated square, and its Ito integral is defined.

Definition: For such a function $f$, write $I(f)=\int_0^T f(t)\,dB_t$ for its deterministic Wiener integral. Functions equal almost everywhere define the same integral. On the real space $L^2([0,T])$, the inner product is $\langle f,g\rangle=\int_0^T f(t)g(t)\,dt$, and the norm satisfies $\|f\|_2^2=\langle f,f\rangle$.

We use the standard mean and isometry properties of the Ito integral: for deterministic $f,g\in L^2([0,T])$, $\mathbb{E}[I(f)]=0$, $\mathbb{E}[I(f)^2]=\int_0^T f(t)^2\,dt$, and $\mathbb{E}[I(f)I(g)]=\int_0^T f(t)g(t)\,dt$. The last identity follows by polarization from the Ito isometry, or by applying the isometry to $f+g$ and subtracting the individual second moments.

The Joint Gaussian Law

Theorem: Let $f_1,\ldots,f_n$ be Borel measurable deterministic functions in $L^2([0,T])$, and set $X_j=I(f_j)$. Then the vector $\mathbf{X}=(X_1,\ldots,X_n)$ is centered multivariate Gaussian, with covariance matrix $C$ given by $$ C_{ij}=\operatorname{Cov}(X_i,X_j)=\int_0^T f_i(t)f_j(t)\,dt,\qquad 1\le i,j\le n. $$ For every $\mathbf{u}=(u_1,\ldots,u_n)\in\mathbb{R}^n$, its characteristic function is $$ \mathbb{E}\!\left[\exp\!\left(i\sum_{j=1}^n u_jX_j\right)\right] =\exp\!\left(-\frac12\sum_{i=1}^n\sum_{j=1}^n u_iC_{ij}u_j\right). $$ The matrix $C$ may be singular; the statement includes that case.

Proof. First take deterministic step functions $f_1,\ldots,f_n$ on a common finite partition $0=t_0<t_1<\cdots<t_m=T$. Each integral is a finite sum of constants times Brownian increments $B_{t_k}-B_{t_{k-1}}$. Those increments are independent centered normal variables with variances $t_k-t_{k-1}$. Consequently every linear combination $\sum_j u_j I(f_j)$ is centered normal. Its variance, by the Ito isometry, is $$ \mathbb{E}\!\left[\left(\sum_{j=1}^n u_jI(f_j)\right)^2\right] =\int_0^T\left(\sum_{j=1}^n u_jf_j(t)\right)^2dt =\sum_{i=1}^n\sum_{j=1}^n u_i u_j\int_0^T f_i(t)f_j(t)\,dt. $$ The same formula holds if the variance is zero: the linear combination then has second moment zero and is zero almost surely. Thus the joint characteristic function for step functions has the asserted form.

For general $f_j\in L^2([0,T])$, use the standard density of deterministic step functions in $L^2([0,T])$ to choose step functions $f_{j,k}$ such that $\|f_{j,k}-f_j\|_2\to0$ as $k\to\infty$. Let $X_{j,k}=I(f_{j,k})$. The Ito isometry gives $\mathbb{E}[(X_{j,k}-X_j)^2]=\|f_{j,k}-f_j\|_2^2\to0$. For fixed $\mathbf{u}$, the Cauchy–Schwarz inequality therefore gives $$ \mathbb{E}\left|\sum_{j=1}^n u_j(X_{j,k}-X_j)\right| \le \sum_{j=1}^n |u_j|\,\|f_{j,k}-f_j\|_2\longrightarrow0. $$ Using $|e^{ix}-e^{iy}|\le |x-y|$, the characteristic functions of these linear combinations converge to that of $\sum_j u_jX_j$. Their variances converge to $\|\sum_j u_j f_j\|_2^2$, since $\|\sum_j u_j(f_{j,k}-f_j)\|_2\le\sum_j|u_j|\|f_{j,k}-f_j\|_2\to0$. Taking the limit in the step-function characteristic-function formula yields the displayed formula for $\mathbf{X}$. The matrix entries are its covariances by the Ito isometry and zero means. The standard uniqueness theorem for characteristic functions identifies the joint law; the approximation and limiting argument above establish the formula without requiring $C$ to be invertible. $\square$

The matrix $C$ is a Gram matrix: its entries are inner products of the integrands. In particular, for every $\mathbf{u}\in\mathbb{R}^n$, $$ \mathbf{u}^{\mathsf T}C\mathbf{u} =\int_0^T\left(\sum_{j=1}^n u_j f_j(t)\right)^2dt\ge0. $$ So $C$ is positive semidefinite. If it is positive definite, the joint law has the usual Gaussian density on $\mathbb{R}^n$. If it is singular, the law is still Gaussian but is concentrated on a lower-dimensional subspace. Equivalently, in either case, $\mathbf{X}$ has the same law as $C^{1/2}\mathbf{Z}$, where $\mathbf{Z}$ has independent standard normal components and $C^{1/2}$ is the positive semidefinite square root of $C$.

Orthogonality and Independence

Corollary: For a finite family of deterministic Wiener integrals, pairwise orthogonal integrands give independent integrals. More generally, two subvectors of a jointly Gaussian vector are independent if every covariance between a component of the first and a component of the second is zero.

Proof. Consider two subvectors and write the joint characteristic function using the theorem. If every cross-covariance is zero, the quadratic expression in its exponent separates into the sum of one expression involving only the first subvector and one involving only the second. Hence the joint characteristic function is the product of the two marginal characteristic functions. By the uniqueness theorem for characteristic functions, the joint law is the product of the marginal laws, which is independence. Pairwise orthogonality is the special case in which the subvectors each have one component. $\square$

Worked Example: Overlapping Integrands

On $[0,2]$, take $f_1=\mathbf{1}_{[0,1]}$ and $f_2=\mathbf{1}_{[0,1]}+2\mathbf{1}_{(1,2]}$. Find the joint law of $X=I(f_1)$ and $Y=I(f_2)$.

The variance of $X$ is the integral of $f_1^2$: it equals $1\cdot1=1$, since $f_1=1$ on an interval of length $1$ and is zero elsewhere. The covariance is $$ \operatorname{Cov}(X,Y)=\int_0^1 1\cdot1\,dt+\int_1^2 0\cdot2\,dt=1+0=1. $$ For $Y$, its variance is $$ \operatorname{Var}(Y)=\int_0^1 1^2\,dt+\int_1^2 2^2\,dt=1+4=5. $$ Thus the mean is $(0,0)$ and the covariance matrix is $C=\begin{pmatrix}1&1\\1&5\end{pmatrix}$. To see the joint structure directly, let $U=B_1-B_0$ and $V=B_2-B_1$. These are independent $N(0,1)$ variables, and the step-integral definition gives $X=U$ and $Y=U+2V$. This representation verifies the variances and covariance: $\operatorname{Var}(X)=1$, $\operatorname{Cov}(X,Y)=1$, and $\operatorname{Var}(Y)=1+4=5$. Also $\det C=1\cdot5-1\cdot1=4>0$, so the joint Gaussian law is nondegenerate.

Worked Example: Orthogonal Integrands Give Independent Integrals

On $[0,2]$, take $f_1=\mathbf{1}_{[0,1]}$ and $f_2=\mathbf{1}_{(1,2]}$. Compute the covariance matrix and identify the joint law.

Both functions have squared integral $1$, because each equals $1$ on an interval of length $1$. Their product is zero except possibly at an endpoint, which does not affect a Lebesgue integral, so $\int_0^2 f_1(t)f_2(t)\,dt=0$. Therefore $C=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. The integrals are $X=B_1-B_0$ and $Y=B_2-B_1$. These are independent $N(0,1)$ increments, in agreement with the orthogonality corollary. The joint characteristic function is $\exp(-(u_1^2+u_2^2)/2)$, which factors into the two standard normal characteristic functions.

Check: the covariance calculation gives zero, while each diagonal entry is the length of its interval, namely $1$. Thus this example has independent components, not merely components with equal variances.

Worked Example: Negative Covariance

On $[0,2]$, let $f_1=\mathbf{1}_{[0,1]}$ and $f_2=-\mathbf{1}_{[0,1]}+\mathbf{1}_{(1,2]}$. Find the covariance matrix and a representation of the pair.

The first variance is $\int_0^1 1\,dt=1$. The covariance is $\int_0^1(1)(-1)\,dt+\int_1^2(0)(1)\,dt=-1+0=-1$. The second variance is $\int_0^1(-1)^2\,dt+\int_1^2 1^2\,dt=1+1=2$. Hence $$ C=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}. $$ With the independent $N(0,1)$ increments $U=B_1-B_0$ and $V=B_2-B_1$, the integrals are $X=U$ and $Y=-U+V$. This confirms $\operatorname{Cov}(X,Y)=-1$, because $\mathbb{E}[U(-U+V)]=-\mathbb{E}[U^2]+\mathbb{E}[UV]=-1+0=-1$. The determinant is $1\cdot2-(-1)(-1)=1>0$, so this joint law is also nondegenerate.

Worked Example: A Singular Joint Gaussian Law

On $[0,1]$, let $f_1(t)=1$ and $f_2(t)=2$. Find the law of $(X,Y)=(I(f_1),I(f_2))$.

The variances are $\int_0^1 1^2\,dt=1$ and $\int_0^1 2^2\,dt=4$, while the covariance is $\int_0^1 1\cdot2\,dt=2$. Thus $C=\begin{pmatrix}1&2\\2&4\end{pmatrix}$, whose determinant is $1\cdot4-2\cdot2=0$. In fact, linearity of the integral gives $Y=2X$, and $X=B_1-B_0\sim N(0,1)$. Therefore the pair has the law of $(Z,2Z)$ for $Z\sim N(0,1)$; it is concentrated on the line $y=2x$ and has no density on the full plane.

Check: the difference $Y-2X$ has variance $4+4(1)-4(2)=0$, so it is zero almost surely. This verifies the stated linear relation from the covariance matrix as well.

How to Read the Covariance Matrix

For a finite collection of deterministic integrands, each entry can be computed directly, without first finding a representation of the integrals as Brownian increments. The diagonal entries are squared $L^2$ norms, so they are nonnegative variances. An off-diagonal entry may be positive, zero or negative according to the signed overlap of the functions. Cauchy–Schwarz also gives the useful check $$ |C_{ij}|=\left|\int_0^T f_i(t)f_j(t)\,dt\right| \le \left(\int_0^T f_i(t)^2\,dt\right)^{1/2} \left(\int_0^T f_j(t)^2\,dt\right)^{1/2}. $$ A proposed covariance matrix that violates this bound, or is not positive semidefinite, cannot be correct for these integrals.

Common mistake: treating integrals with different-looking integrands as independent without checking their inner product. For example, on $[0,1]$, $I(1)$ and $I(-1)$ have different integrands, but their covariance is $\int_0^1(1)(-1)\,dt=-1$, not zero. In fact $I(-1)=-I(1)$ exactly. Independence requires orthogonality for these jointly Gaussian integrals; visual difference between the integrands is not enough.

The deterministic assumption matters. The Gaussian conclusion proved here concerns integrands fixed in advance, not integrands chosen using the observed Brownian path. In this setting, the entire finite-dimensional law is encoded by the deterministic $L^2$ geometry of the functions: inner products give covariances, and zero cross-inner-products give independence.

Check Your Understanding

Use the covariance formula and the joint Gaussian result to answer each question. Treat integrands that differ only at endpoints as equal in $L^2$.

On $[0,3]$, find the covariance of $I(\mathbf{1}_{[0,2]})$ and $I(\mathbf{1}_{[1,3]})$, and find both variances.
For two deterministic $L^2$ integrands with inner product zero, what additional fact about their joint law lets you conclude that their integrals are independent?
Suppose $g=-3f$ and $\int_0^T f(t)^2\,dt=2$. Find the covariance matrix of $(I(f),I(g))$ and state the almost-sure relation between its components.
Explain why a zero determinant of the covariance matrix does not prevent a pair of deterministic Wiener integrals from having a well-defined joint law.
Give an inequality that every covariance entry must satisfy in terms of the corresponding variances.